Question

x^2/(x+1)-ax-b当n->无穷时的极限为0,求a,b

Answer 1

x^2/(x+1)-ax-b =(x^2-(x+1)(ax+b))/(x+1) =((1-a)x^2-(a+b)x-b)/(x+1) 当x趋于无穷时,极限为0 则1-a=0,a+b=0 故a=1,b=-1